Hi Bert, I agree that our interpretation is slightly different, which is why I wrote: "If one can take the actual week of the year as an acceptable definition of week, then here's my approach."
Regards, Eric On Wed, Jun 19, 2019 at 5:04 PM Bert Gunter <bgunter.4...@gmail.com> wrote: > Eric: > > I believe you're doing something different than I did. I broke up each > month into biweekly periods, 2+ per month. You seem to be grouping the > overall entire period into biweekly intervals -- apologies if I'm wrong, > but if I understood correctly, that's not the same thing. I do not know > which of us -- if either -- has interpreted her query correctly. > > Cheers, > Bert > > On Wed, Jun 19, 2019 at 2:35 AM Eric Berger <ericjber...@gmail.com> wrote: > >> Hi Siti, >> I didn't test Bert's code but I assume it's fine. :-) >> I would take a different approach than Bert. I prefer to use a package >> such as lubridate to handle the date wrangling, and a package such as dplyr >> to handle the grouping and max extraction. >> It may be overkill for this problem, but these are great packages to >> become familiar with. >> If one can take the actual week of the year as an acceptable definition >> of week, then here's my approach. >> >> library(lubridate) >> library(dplyr) >> >> # Step 1: start with Bert's code to create sample data >> ## create some example data for 3 months in 2000 >> d<- 2e7 +c(113:131,201:228, 301:330) ## dates >> conc <- runif(length(d)) # concentrations >> >> # Step 2: collect the data into a data frame >> df <- data.frame( dt=d, conc=conc) >> >> # Step 3: use lubridate's ymd() function to parse the dates, its week() >> function to identify the week of the year, and define the new column >> 'wkpair' that groups the weeks 2-at-a-time >> df2 <- dplyr::mutate( df, >> wkpair=as.integer(floor(lubridate::week(lubridate::ymd(dt) )/2)) ) >> >> # Step 4: group by the wkpair and use dplyr's summarise to get the info >> you wanted >> df3 <- dplyr::group_by(df2,wkpair) %>% >> dplyr::summarise( from=min(dt), to=max(dt), maxconc=max(conc)) >> %>% >> dplyr::select(from,to,maxconc) >> >> df3 >> >> # A tibble: 6 x 3 >> from to maxconc >> <dbl> <dbl> <dbl> >> 1 20000113 20000121 0.963 >> 2 20000122 20000204 0.988 >> 3 20000205 20000218 0.939 >> 4 20000219 20000303 0.883 >> 5 20000304 20000317 0.863 >> 6 20000318 20000330 0.765 >> >> HTH, >> Eric >> >> >> >> On Tue, Jun 18, 2019 at 9:39 PM Bert Gunter <bgunter.4...@gmail.com> >> wrote: >> >>> My apologies. I negected to cc r-help. -- Bert >>> >>> >>> >>> On Tue, Jun 18, 2019 at 11:21 AM Bert Gunter <bgunter.4...@gmail.com> >>> wrote: >>> >>> > >>> > I assume that 20000215 means year 2000, month 2, day 15. >>> > I also assume that you want maxes for the first 2 weeks of a month, the >>> > second 2 weeks, and any remaining days. >>> > I also assume that this might be desired for arbitrary numbers of >>> years, >>> > months, and days. >>> > >>> > The following is one way to do this. As it's a slight pain to cut and >>> > paste email data as text into R (use ?dput or R code to run to provide >>> > example data instead), I just made up my own. You'll have to do the >>> > following within a data frame through extraction or by using with() of >>> > course. >>> > >>> > ## create some example data for 3 months in 2000 >>> > d<- 2e7 +c(113:131,201:228, 301:330) ## dates >>> > conc <- runif(length(d)) # concentrations >>> > >>> > ## convert the date to character to extract year, month, and day >>> > cdate <- as.character(d) >>> > ## use substr to to the extraction >>> > year <- substr(cdate,1,4) >>> > mon <- substr(cdate,5,6) >>> > day <- substr(cdate, 7,8) >>> > >>> > ## convert day to numeric and use cut() to group into the biweekly >>> periods. >>> > d14 <- cut(as.numeric(day), c(0,14.5,28.5, 32)) >>> > >>> > ## Use tapply() to create your desired table of results. >>> > tapply(conc, list(year, d14, mon), max, na.rm = TRUE) >>> > >>> > ## Results >>> > >>> > , , 01 >>> > >>> > (0,14.5] (14.5,28.5] (28.5,32] >>> > 2000 0.7357389 0.9655391 0.7962965 >>> > >>> > , , 02 >>> > >>> > (0,14.5] (14.5,28.5] (28.5,32] >>> > 2000 0.8193979 0.9487207 NA >>> > >>> > , , 03 >>> > >>> > (0,14.5] (14.5,28.5] (28.5,32] >>> > 2000 0.9718919 0.9997093 0.168659 >>> > >>> > >>> > Cheers, >>> > Bert >>> > >>> > Bert Gunter >>> > >>> > >>> > >>> > >>> > On Tue, Jun 18, 2019 at 8:53 AM SITI AISYAH BINTI ZAKARIA < >>> > aisyahzaka...@unimap.edu.my> wrote: >>> > >>> >> Hi, >>> >> >>> >> I'm Aisyah..I have a problem to run my R coding. I want to select >>> maximum >>> >> value according to week. >>> >> >>> >> here is my data >>> >> >>> >> Date O3_Conc >>> >> 20000101 0.033 >>> >> 20000102 0.023 >>> >> 20000103 0.025 >>> >> 20000104 0.041 >>> >> 20000105 0.063 >>> >> 20000106 0.028 >>> >> 20000107 0.068 >>> >> 20000108 0.048 >>> >> 20000109 0.037 >>> >> 20000110 0.042 >>> >> 20000111 0.027 >>> >> 20000112 0.035 >>> >> 20000113 0.063 >>> >> 20000114 0.035 >>> >> 20000115 0.042 >>> >> 20000116 0.028 >>> >> >>> >> I want to find the max value from column O3_Conc for only 14 days that >>> >> refer to biweekly in month. And the next 14 days for the max value. >>> >> >>> >> I hope that I can get the result like this: >>> >> >>> >> Date Max O3_Conc >>> >> 20000101 - 20000114 0.068 >>> >> 20000115 - 20000129 0.061 >>> >> >>> >> I try many coding but still unavailable. >>> >> >>> >> this example my coding >>> >> >>> >> library(plyr) >>> >> data.frame(CA0003) >>> >> >>> >> # format weeks as per requirement (replace "00" with "52" and >>> >> adjust corresponding year) >>> >> tmp <- list() >>> >> tmp$y <- format(df$Date, format="%Y") >>> >> tmp$w <- format(df$Date, format="%U") >>> >> tmp$y[tmp$w=="00"] <- >>> as.character(as.numeric(tmp$y[tmp$w=="00"]) - >>> >> 14) >>> >> tmp$w[tmp$w=="00"] <- "884" >>> >> df$week <- paste(tmp$y, tmp$w, sep = "-") >>> >> >>> >> # get summary >>> >> df2 <- ddply(df, .(week),transform, O3_Conc=max(O3_Conc)) >>> >> >>> >> # include week ending date >>> >> tmp$week.ending <- lapply(df2$week, function(x) rev(df[df$week >>> ==x, >>> >> "O3_Conc"])[[1]]) >>> >> df2$week.ending <- sapply(tmp$week.ending, max(O3_Conc, TRUE) >>> >> >>> >> output >>> >> Site_Id Site_Location >>> Date >>> >> Year O3_Conc Month Day week >>> >> 1 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000101 >>> >> 2000 0.033 1 1 NULL-NULL >>> >> 2 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000102 >>> >> 2000 0.023 1 2 NULL-NULL >>> >> 3 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000103 >>> >> 2000 0.025 1 3 NULL-NULL >>> >> 4 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000104 >>> >> 2000 0.041 1 4 NULL-NULL >>> >> 5 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000105 >>> >> 2000 0.063 1 5 NULL-NULL >>> >> 6 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000106 >>> >> 2000 0.028 1 6 NULL-NULL >>> >> 7 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000107 >>> >> 2000 0.068 1 7 NULL-NULL >>> >> 8 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000108 >>> >> 2000 0.048 1 8 NULL-NULL >>> >> 9 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000109 >>> >> 2000 0.037 1 9 NULL-NULL >>> >> 10 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000110 >>> >> 2000 0.042 1 10 NULL-NULL >>> >> 11 CA0003 Sek. Keb. Cederawasih, Taman Inderawasih, Perai >>> 20000111 >>> >> 2000 0.027 1 11 NULL-NULL >>> >> >>> >> >>> >> >>> >> >>> >> -- >>> >> This message has been scanned by E.F.A. 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